# Can one describe the algebraic multiplicity in terms of generalized eigenspaces and the minimal polynomial?

We defined the algebraic multiplicity of a matrix $A$ with eigenvalue $\lambda$ to be the largest integer $r$ such that $(x-\lambda)^r$ divides the characteristic polynomial of $A$.

I would like to know if that is the same as the minimal integer $s$ so that the kernel of $(A-\lambda I)^s$ is the same as the kernel of $(A-\lambda I)^{s+1}$. I am not really sure if that is true and how one can prove that. And does $(x-\lambda)^s$ divide the minimal polynomial then?

Could you clear up this question for me? If possible, is there a proof for this claim without using the Jordan Normal Form? I know about the decomposition in (invariant) generalized eigenspaces though.

Thank you!

• No, this is not true. For $A = I_n$, the only eigenvalue is $\lambda = 1$, and the characteristic polynomial is $(x - 1)^n$, so the algebraic multiplicity is $n$, but $A - I_n = 0$, so $\ker (A - I_n) = \ker ((A - I_n)^2)$, and hence the minimal integer $s$ is $1$. – Travis Jun 27 '18 at 2:52
• It is true that $(x - \lambda)^s$ divides the minimal polynomial of $A$, and is in fact the largest power of $(x - \lambda)$ which divides the minimal polynomial. In terms of the Jordan form, $s$ is the size of the largest Jordan block associated with $\lambda$. – Omnomnomnom Jun 27 '18 at 6:33

The minimal integer $s$ such that $\ker (A -\lambda I)^s = \ker (A - \lambda I)^{s+1}$ is precisely the largest power of $x-\lambda$ which divides the minimal polynomial of $A$. This is in general $\le$ than the algebraic multiplicity of $\lambda$.

Recall that in general holds $\ker T \subseteq \ker T^2 \subseteq \ker T^3 \subseteq \cdots$ and that if $\ker T^m = \ker T^{m+1}$ then $\ker T^m = \ker T^{n}, \forall n \ge m$.

The rest follows from this lemma:

Let $A : V \to V$ be a linear map on a finite-dimensional complex vector space $V$ with the eigenvalues $\{\lambda_1, \ldots, \lambda_r\}$. Then a polynomial $p(x) = (x-\lambda_1)^{m_1}\cdots (x-\lambda_r)^{m_r}$ annihilates the map $A$ if and only if $$V = \ker (A-\lambda_1 I)^{m_1} \,\dot+\,\cdots \,\dot+\,\ker(A-\lambda_r I)^{m_r}$$

The direction $\impliedby$ is more-less obvious, and for $\implies$ you can use the partial fraction decomposition of $\frac1{p(x)}$. Note that this sum is always direct, i.e. we have $\ker (A-\lambda_i I)^{m_i} \cap \ker (A-\lambda_j I)^{m_j} = \{0\}, \forall i \ne j$ so you just need to prove that the sum is equal to $V$.

Let $\mu_A(x) = (x-\lambda_1)^{s_1}\cdots (x-\lambda_r)^{s_r}$ be the minimal polynomial of $A$. Then by the lemma we conclude $$V = \ker (A-\lambda_1 I)^{s_1} \,\dot+\,\cdots \,\dot+\,\ker(A-\lambda_r I)^{s_r}$$

Let $m_i\ge s_i$. Then clearly $(x-\lambda_1)^{m_1}\cdots (x-\lambda_r)^{m_r}$ also annihilates $A$ so $$V = \ker (A-\lambda_1 I)^{m_1} \,\dot+\,\cdots \,\dot+\,\ker(A-\lambda_r I)^{m_r}$$

Now $$\sum_{i=1}^r \dim\ker (A-\lambda_i I)^{m_i} = \dim V = \sum_{i=1}^r \dim\ker (A-\lambda_i I)^{s_i}$$ implies $\dim\ker (A-\lambda_i I)^{m_i} = \dim\ker (A-\lambda_i I)^{s_i}$ so $\ker (A-\lambda_i I)^{m_i} = \ker (A-\lambda_i I)^{s_i}$ for all $i = 1,\ldots, r$.

On the other hand, if $k_i \le s_i$ and we assume $\ker (A - \lambda I)^{k_i} = \ker (A - \lambda I)^{s_i}$, then we have

$$V = \ker (A-\lambda_1 I)^{k_1} \,\dot+\,\cdots \,\dot+\,\ker(A-\lambda_r I)^{k_r}$$

so the lemma implies that the polynomial $p(x) = (x-\lambda_1)^{k_1}\cdots (x-\lambda_r)^{k_r}$ annihilates $A$. Hence the minimal polynomial divides $p$ so $k_i = s_i$ for all $i = 1, \dots, n$.